Proving Thomae's Function f(x) Not Differentiable"

[limdaesung]

Thomae function f(x):(0,1)->R f(x)=p/q x is rational number (p and q are relatively prime natural number) f(x)=0 x in irrational number show that f is not differrentiable. l can show that this function is not differentiable at rational number. But i can't sequence that is example. not differrentiable at irrational number. how take this sequence.

 

[limdaesung]

please i want to find this sequence quickly.

 

[HallsofIvy]

A good start would be rereading the problem and then writing it down correctly! What you have written makes no sense.

Thomae function f(x):(0,1)->R f(x)=p/q x is rational number (p and q are relatively prime natural number) f(x)=0 x in irrational number

I presume that means that if is an irrational number then f(x)= 0, but it's not clear to me what f(x) is if x is rational. You say f(x)= p/q but you haven't told us what numbers p and q are.

I might be inclined to guess that you mean "if x is a rational number, x= p/q with p and q relatively prime, the f(x)= 1/q." That's a fairly well known function, often called the "modified Dirichlet function". (The regular Dirichlet function is f(x)= 0 if x is irrational, f(x)= 1 if x is rational.)

If that is the function you are talking about, it can be shown that lim(x->a) f(x)= 0 for all a so the modified Dirichlet function is continuous for all irrational a, discontinuous for rational a. Since a function cannot be differentiable where it is not continuous, if this is the right function, then I presume the problem is to show that this function, even though it is continuous at each irrational number, is not differentiable there.